To determine the empirical formula of a compound given that it contains 43.6% oxygen by mass, follow these steps:
1. Calculate the percentage of phosphorus in the compound:
Since the compound contains 43.6% oxygen, the remaining percentage must be phosphorus:
[tex]\[
100\% - 43.6\% = 56.4\%
\][/tex]
2. Set up the ratio of oxygen to phosphorus by mass:
The molar mass of oxygen (O) is 16 g/mol, and the molar mass of phosphorus (P) is 30.97 g/mol.
3. Calculate the mass-to-mole ratio for each element:
- For phosphorus:
[tex]\[
\text{Moles of phosphorus} = \frac{56.4}{30.97} \approx 1.821
\][/tex]
- For oxygen:
[tex]\[
\text{Moles of oxygen} = \frac{43.6}{16} \approx 2.725
\][/tex]
4. Determine the simplest mole ratio between the elements:
We calculate the ratio of moles of oxygen to moles of phosphorus:
[tex]\[
\text{Oxygen to Phosphorus ratio} = \frac{2.725}{1.821} \approx 1.496
\][/tex]
5. Convert the mole ratio to a small whole number ratio:
The ratio 1.496 is approximately 1.5 suggesting that for every phosphorus (P) atom, there are roughly one and a half oxygen (O) atoms. To express this as a simple whole number ratio, we multiply both sides by 2:
[tex]\[
1.5 \text{ (O)} \approx 3 \text{ (P)} \rightarrow 3 \times 2 \approx 2 \text{ (O)} \times 5
\rightarrow \text {P}_2\text{O}_5
\][/tex]
Hence, the empirical formula of the compound is:
[tex]\[
\boxed{P_2O_5}
\][/tex]
